today in physics 122: dc circuits - university of rochesterdmw/phy122/lectures/lect_16b.pdf ·...
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Today in Physics 122: DC circuits
Currents and voltages in purely series/parallel resistor-battery circuits
Conservation of charge and energy, and Kirchhoff’s Rules.
Currents in general resistor-battery circuits
Gustav Kirchhoff, inventor of The Rules.
4 October 2019 Physics 122, Fall 2019 1
DC
DC stands for direct current.
All that means is that the current isn’t alternatingperiodically.
It’s often taken to mean, though that the currents and voltages in a circuit are constant, which in principle would only apply to a circuit with constant voltage sources and resistors.
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DC circuit components and their voltages
As usual in circuits, we will speak loosely about potential V and potential difference ∆V, normally just using V.
Resistor: V = IR
Capacitor: V = Q/C
Perfect conductor: V = 0
DC power source: V = constant
One paradigm of the DC power source is the battery, in which chemical reactionsgenerate an electromotive force (EMF)
which allows the device to draw a current without changing its voltage, at least by very much.
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Alessandro Volta, who invented the battery in 1800, and for whom the Volt is named.
Reminder: combinations of Rs, Cs
Parallel and series pairs of resistors and capacitors:
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1C
2C
1 2
1 2
1 2
1 2
1 1 1
,
eq
eq
C C C
C CCC CC C
= +
=+
<
1C 2C1 2
1 2,eqC C C
C C
= +
>
1R
2R
1 2
1 2
1 2
1 2
1 1 1
,
eq
eq
R R R
R RRR RR R
= +
=+
<
1R 2R
1 2
1 2,eqR R R
R R
= +
>
=
=
=
=
Currents and voltages in DC circuits
Physicists and engineers often need to construct electrical circuits, and to specify the currents and voltages of each component.
This you already know how to do for networks that can be decomposed into purely series and parallel combinations of resistors and capacitors.
For example, the resistor pyramid (30 September): let’s calculate the current in each resistor. All resistors have the same value, R.
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V
Currents in the resistor pyramid
We showed before (30 September) that the pyramid is equivalent to these resistor combinations:
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=2R
3R
R
4R
= 2512
R
Currents in the resistor pyramid (continued)
The total current drawn I from the battery is obtained from the fact that the battery voltage V is applied to the resistance equivalent to the whole pyramid:
4 October 2019 Physics 122, Fall 2019 7
2512
R
V
I
1225eq
V VIR R
= =
Currents in the resistor pyramid (continued)
This same current is carried by each of the series of resistors that the pyramid is also equivalent to. By Ohm’s Law their voltages are respectively
4 October 2019 Physics 122, Fall 2019 8
V
2R
3R
R
4R
1
2
3
4
12256
2 254
3 253
4 25
VV IR
IR VV
IR VV
IR VV
= =
= =
= =
= = 1225
VIR
=
Currents in the resistor pyramid (continued)
These voltages can be applied to each of the parallel combinations in the pyramid:
Because the resistors are all the same, the currents in each resistor in the parallel combos is the same.
4 October 2019 Physics 122, Fall 2019 9
V
625V
425V
325V
1225
V
11
22
33
44
12 ( )25625425325
V VI IR RV VIR R
V VIR R
V VIR R
= = =
= =
= =
= =
1I
2I 2I
3I 3I 3I
4I 4I 4I 4I
Resistor-battery circuits in general
All that is OK, but it isn’t hard to conceive circuits in which the resistors are neither in series nor in parallel with each other. For this we need to elaborate on another couple of electrostatics facts:
Conservation of electric charge
Conservation of energy
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1V
2V
2R
1R
Kirchhoff’s Rules
Electric charge is conserved: it can’t be created or destroyed.
Nor can charge accumulate in a perfectly-conducting junction between circuit components, or in a resistor.
Thus at each node – the conductor of whatever shape that joins two or more components –the current flowing in is equal to the current flowing out.
This is called Kirchhoff Rule #1.
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1I 3I 5I
2I 4I
1 2 3 4 5 0I I I I I− + − + =
Node
Kirchhoff’s Rules (continued)
Energy is conserved, and potential difference is work per unit charge.
Therefore the total work done on a charge that is transported all the way through a continuous path through a circuit, back to its starting point, must be zero.
And therefore the voltages it traverses on the way through any loop in the circuit must add up to zero.
This is called Kirchhoff Rule #2.
4 October 2019 Physics 122, Fall 2019 12
1I
3I
1R
2I
4I
4R
3R
2R
V
1 1 2 2
3 3 4 4 0V I R I R
I R I R− ++ + =
Loop
Kirchhoff’s Rules (continued)
To implement Kirchhoff’s Rules on a circuit:
Define a current – value and direction –through each component in the circuit.
In the node rule (KR#1), bookkeep all currents flowing into the node as positive, and all those flowing out as negative; the sum of all the positives and negatives comes to zero.
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1I 3I 5I
2I 4I
1 2 3 4 5 0I I I I I− + − + =
Kirchhoff’s Rules (continued)
Current flows from the high (+) to the low (-) potential end of each component.
In the loop rule (KR#2), count voltages traversed from - to + as positive, and those traversed from + to – as negative.
• That is, voltage dropsacross a resistor if you’re following voltage in the direction the current flows through it.
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1I
3I
1R
2I
4I
4R
3R
2R
V
1 1 2 2
3 3 4 4 0V I R I R
I R I R− ++ + =
++
+
+
+
-
--
-
-
Kirchhoff’s Rules (continued)
Identify the unknown quantities – N, say – in the circuit, and count them.
Then write the node rule and/or the loop rule to generate as many relations between the voltages and currents as there are unknowns (N).
• Use both the node rule and the loop rule, at least once each.
This gives a system of N equations in N unknowns, which is an algebra problem you first learned how to solve in middle school. (No calculus required! Yet.)
Important point: it doesn’t matter whether you correctly guess the direction of each current. If you guess wrong, your answer will just be a negative number. You will still know which way the current flows.
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Example: two resistors, two batteries
In the circuit shown here,
Find the current in each resistor if
and vice versa:
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1V
2V2R
1R
1
2
100200
RR
= Ω= Ω
1
2
10 V1 V,
VV
==
1
2
1 V10 V.
VV
==
Two resistors, two batteries (continued)
Identify unknowns: There are three branches to the circuit, and so there are three currents. We don’t know any of them. Those are our unknowns; they are identified at right.
There are two nodes, at top and bottom. Write the node equation for the top one :
4 October 2019 Physics 122, Fall 2019 17
1V
2V
2R
1R
1I
2I
1 2 3 0I I I− + =
Node
Node
3I
Two resistors, two batteries (continued)
There are three loops. Write the loop equation for two of them: we’ll pick the two indicated and traverse them clockwise starting at lower left.
Now solve. The first loop equation gives us I3 right away:
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++
+ +-
-
-
-1 3 1 2
2 3 1 2 2
00
V I R VV I R I R
+ − =− − =
2 13
1
V VIR−
=
1V
2V
2R
1R
3I
1I
2I
Two resistors, two batteries (continued)
And I3 can be put right back into the other loop equation to find I2 :
Putting these two into the node equation gives the remaining current:
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( )
2 3 12
2
2 2 1 1
2 2
V I RI
RV V V V
R R
−=
− −= =
2 1 11 3 2
1 2
V V VI I IR R−
= − = −
++
+ +-
-
-
-1V
2V
2R
1R
3I
1I
2I
Two resistors, two batteries (continued)
Now, and only now, we put the numbers in:
For
4 October 2019 Physics 122, Fall 2019 20
++
+ +-
-
-
-1
22
2 13
1
0.05 A
0.09 A
VIRV VI
R
= =
−= = −
2 30.005 A, 0.09 AI I= = +
1 2
1 2
100 , 20010 V, 1 V:
R RV V
= Ω = Ω= =
1 21 V, 10 V:V V= =
1V
2V
2R
1R
3I
1I
2I
Two resistors, two batteries (continued)
Notes on the solution:
Intuitively obvious, perhaps, since V1 is in parallel with R2.
Note that the two node equations are not independent:
In general, if there are N nodes, only N-1 of the node equations are independent, which is why we need loop equations in addition.
4 October 2019 Physics 122, Fall 2019 21
++
+ +-
-
-
-1 2 3
2 1 3
00
I I II I I− + =− − =
1V
2V
2R
1R
3I
1I
2I
Two resistors, two batteries (continued)
The same with the loop equations: here there are three, but only two are independent. In general if there are M, M-1 are independent.
But between node and loop equations there are alwaysenough independent equations to solve for the unknowns.
4 October 2019 Physics 122, Fall 2019 22
++
+ +-
-
-
-1V
2V
2R
1R
3I
1I
2I